Simplify and expand the following expression: $ \dfrac{6}{5t - 6}-\dfrac{t - 2}{2t - 8} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(5t - 6)(2t - 8)$ Multiply the first term by $\dfrac{2t - 8}{2t - 8}$ $ \begin{align*} \dfrac{6}{5t - 6} \times \dfrac{2t - 8}{2t - 8} & = \dfrac{(6)(2t - 8)}{(5t - 6)(2t - 8)} \\ & = \dfrac{12t - 48}{(5t - 6)(2t - 8)}\end{align*} $ Multiply the second term by $\dfrac{5t - 6}{5t - 6}$ $ \begin{align*} \dfrac{t - 2}{2t - 8} \times \dfrac{5t - 6}{5t - 6} & = \dfrac{(t - 2)(5t - 6)}{(2t - 8)(5t - 6)} \\ & = \dfrac{5t^2 - 16t + 12}{(2t - 8)(5t - 6)}\end{align*} $ Now we have: $ = \dfrac{12t - 48}{(5t - 6)(2t - 8)} - \dfrac{5t^2 - 16t + 12}{(2t - 8)(5t - 6)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{12t - 48 - (5t^2 - 16t + 12)}{(5t - 6)(2t - 8)} $ $ = \dfrac{12t - 48 - 5t^2 + 16t - 12}{(5t - 6)(2t - 8)} $ $ = \dfrac{28t - 60 - 5t^2}{(5t - 6)(2t - 8)}$ Expand the denominator: $ = \dfrac{28t - 60 - 5t^2}{10t^2 - 52t + 48}$